totem pole output

Discussion in 'Homework Help' started by electronicsbeginner12, Nov 27, 2015.

  1. electronicsbeginner12

    Thread Starter New Member

    Dec 14, 2014
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    In my text, it describes the time it takes for the output to transition from high to low is dependent on the rc time constant of R2 and the gate capacitance as shown in the attached picture. However, there is R3 in the picture that is said to be in parallel with R2, and since R2 is so much smaller than R3, the combo will have a value approximately of R2. Im confused as to why R3 is parallel with R2. Can someone give me some insights. Thanks in advance.

    20151127_024529_695x1024.jpg
     
  2. Alec_t

    AAC Fanatic!

    Sep 17, 2013
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    For a properly decoupled power supply, Vcc is effectively at ground potential as far as AC is concerned. Therefore, when Q2 is closed, R3 appears in parallel with R2.
     
  3. electronicsbeginner12

    Thread Starter New Member

    Dec 14, 2014
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    My text makes no mention of this concept. When you say Vcc is effectively an AC short, does it mean all of the AC signal goes through Vcc is at ground potential?

    Also, if the RC discharge is DC current, why would i need to consider the ac current that is going to Vcc through R3?

    Im trying to understand this on a basic level. Thanks for your help.
     
  4. Alec_t

    AAC Fanatic!

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    Not sure what you mean.
    When C is discharging through R2 and Q2, there is a small current flowing through R3; but since R3 is 20 times bigger than R2 the R3 current is negligible by comparison with the R2 current.
    Can you quote the relevant text section, or post a link to it?
     
  5. electronicsbeginner12

    Thread Starter New Member

    Dec 14, 2014
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    I have attached an image of the text section. I dont understand how I need to consider that Vcc is considered an AC ground and that R3 is effectively in parallel to R2. The concept of Vcc being an AC ground isn't explained in my text and is rather confusing. Do I just treat every Vcc as ground for simplicity to calculate the equivalent resistance? Thanks.
     
    Last edited: Nov 28, 2015
  6. bertus

    Administrator

    Apr 5, 2008
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    Hello,

    Please keep the images posted in the correct orientation.
    I will not turn my monitor to read it.

    Electronicsbeginner12_text.jpg

    I rotated and changed contrast of the image.

    Bertus
     
    absf likes this.
  7. Alec_t

    AAC Fanatic!

    Sep 17, 2013
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    That puts a somewhat different complexion on things. When the cap is charging (Q1 conducting), R3 is in parallel with the series combination of R1 and R2 and would be more significant than when the cap is discharging. The text is misleading. For your purpose you can forget my AC comment.
     
  8. WBahn

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    Mar 31, 2012
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    One common way of analyzing a circuit is to break it's behavior into a large signal response and a small signal response and, using superposition, get the total response as the sum of the two. In applying superposition, you segregate all of the independent signal sources into two groups (in this case), those responsible for the large signal response and those responsible for the small signal response. The terms "large signal" and "small signal" are customary, but it really is nothing more than the response of the circuit to two different groups of sources. The names come from the types of signals we USUALLY work with in each group. For similar reasons, we also call these two groups the DC response and the AC response, but again the names are just names.

    When you apply superposition to one set of signals, you zero out the sources in the other group. Since the Vcc supply is in the large signal (the DC) group, it is set to 0 V when determining the response to the signals in the small signal (the AC) group, which include the switching input signal.

    A more convincing way to see this is to do the full solution to the transient response for when the lower switch is closed and you will see that the result has Vcc involved in determining the final steady state result (this is the large-signal component) while the resistors combine as if they were in parallel for determining the time constant of the transient portion of the response (this is the small-signal component).
     
  9. electronicsbeginner12

    Thread Starter New Member

    Dec 14, 2014
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    Thanks for the insight. However, it is a bit over my head right now. I haven't taken a course on network analysis yet, and I am just trying to understand on a bare basic level at this time to get through my text. My text doesn't explain some concepts before introducing them, which it makes it difficult. For simplicity, can I just consider these resistors to be in parallel while current is discharging through R2 and Q2 because Vcc of Q3 is considered an AC ground? What I don't understand is why is Vcc an AC ground. My text makes no mention of this concept. A simple explanation would be helpful. Thanks.
     
  10. crutschow

    Expert

    Mar 14, 2008
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    Vcc is a DC voltage. A perfect DC voltage is a constant value independent of any current into or out of it (this is achieved in practice by having large decoupling capacitors connected from the voltage to ground).
    Thus, for an AC current, Vcc looks the same as ground.
     
  11. MrAl

    Well-Known Member

    Jun 17, 2014
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    Hi,

    Just as a general note which applies to many circuits...

    Sometimes you are after an *exact* response and sometimes an *approximate* response, however sometimes the meaning of these can vary where sometimes they might even be considered equal to each other.

    But in the context where one resistor affects the response more than another, the exact analysis is to include that second resistor, no matter how small the effect may be, while for the approximate response we may leave that resistor out entirely.

    They both have their purposes. For the approximate response, it works right now, with the actual values we see at the present time. For the exact response, that works right now and even later when we may decide to change that resistor value where it then makes a more significant impact on the total response.

    For resistors that are 100 fold different (like one 1 ohm and one 100 ohms) the 100 ohm may not make much difference, or the 1 ohm resistor may not make much difference. But once we change the 1 ohm to 100 ohms or the 100 ohms to 1 ohm, we could see the response vary by a huge amount. Put the point is, with an exact analysis we still get the right output response, whether we keep the resistors as is or make one larger or smaller which brings it's value closer to the other value.

    In the approximate analytical analysis, we get one result or we approximate for several values.
    In the exact analytical analysis, we get a result which works for many circuits like this one and it works for any value.

    In the exact numerical analysis, we get a single result, but then we can compare it to the approximate numerical analysis so we can determine if the approximate analysis is applicable and possibly the range over which that approximation is valid.

    So there are different ways to analyze depending on what we hope to learn immediately and how applicable we want the result of that analysis to be in the future.

    A more clear example might be like this:
    i=i1*(e^(v/K1)-1)+K2*i1

    which is the exact response, but we may find that a simpler approximation works over a certain range such as:
    i=i1*e^(v/K3)

    and that makes the analysis of the dependence of v on i easier to calculate:
    v=K3*ln(i/i1)

    however the range of i and v is limited.
     
    Last edited: Dec 13, 2015
  12. electronicsbeginner12

    Thread Starter New Member

    Dec 14, 2014
    27
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    I see. This is because the decoupling cap passes ac to ground, right? Thanks.
     
  13. electronicsbeginner12

    Thread Starter New Member

    Dec 14, 2014
    27
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    Thanks for the lengthy explanation. Yes. For me going through an Electronics Technology program, approximate analysis is the emphasis used for troubleshooting purposes. It glosses over more in depth analysis, but I can see where it can be helpful.
     
  14. WBahn

    Moderator

    Mar 31, 2012
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    No, it has to do with superposition, which means that, for a linear circuit (or a circuit operated over a sufficiently small range that it can be adequately approximated by a linear circuit) each (independent) source's contribution to the total response is independent of the contribution of the other sources.

    When you analyze such a circuit, you end up with a response that looks something like this:

    Vout = A·V1 + B·V2 + C·V3

    You can write this as the sum of three different responses:

    Vout = Response_1 + Response_2 + Response_3

    Where

    Response_1 = A·V1
    Response_2 = B·V2
    Response_3 = C·V3

    As you can see, each response is simply the total response but with the other voltage source set equal to zero. A voltage source that is set equal to 0V behaves the same as a short-circuit. Hence the response of the system to V2, for instance, is the same as the response of a circuit in which V1 and V3 are replaces by short circuits. If one side of V1 and V3 are connected to ground, this means that the other side is no effectively connected to ground.
     
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